The triangle of maximum area inscribed in a circle is -

The circle $x^2 + y^2 - 4x - 4y + 4 = 0$ is...

Let $S=0$ be the circle passing through the points $(2,0)$,$(1,-2)$,and $(-1,1)$. Then the point $(1,2)$

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

The straight line $(x - 2) + (y + 3) = 0$ cuts the circle $(x - 2)^2 + (y - 3)^2 = 11$ at

The line $4x - 3y + 15 = 0$ intersects the circle $x^2 + y^2 - 6x - 8y = 0$ at two points $A$ and $B$. The maximum area of $\Delta ABC$,where $C$ is a point on the circumference of the circle,will be - .............. $sq. \ units$.